Codility has a number of lessons online to help candidates prepare for the problems on the site. I figured it might be worthwhile to make a summary of some of the algorithms from the lessons that I more easily forget.

# Prefix (and suffix) sums

Makes it easy to calculate totals for slices of an array. Codility provides the following algorithm:

```
def prefix_sums(A):
n = len(A)
P = [0] * (n + 1)
for k in range(1, n + 1):
P[k] = P[k - 1] + A[k - 1]
return P
```

Note that the first element is `0`

with this algorithm.

The sum of the elements in a slice from `x`

to `y`

can be calculated using `P[y + 1] - P[x]`

.

# Leader of an array

The leader of an array is the element that occurs more than times in the array. Codility provides the following solution for this:

```
def goldenLeader(A):
n = len(A)
size = 0
# keep a 'stack' of leaders to find a potential candidate
for k in xrange(n):
if (size == 0):
size += 1
value = A[k]
else:
if (value != A[k]):
size -= 1
else:
size += 1
# validate that the candidate is actually the leader
candidate = -1
if (size > 0):
candidate = value
leader = -1
count = 0
for k in range(n):
if (A[k] == candidate):
count += 1
if(count > n // 2):
leader = candidate
# returns -1 if no leader was found
return leader
```

# Maximum slice

Finds the slice with the largest sum. Codility provides an implementation of Kadane’s algorithm for this problem:

```
def golden_max_slice(A):
max_ending = max_slice = 0
for a in A:
max_ending = max(0, max_ending + a)
max_slice = max(max_slice, max_ending)
return max_slice
```

Initialisation doesn’t have to start from 0 and Wikipedia has more information on variations of the algorithm.

# Counting divisors of

Codility provides an algorithm that can count divisors of in . It relies on finding symmetric divisors, which allows you to count two divisors for the price of one.

```
def divisors(n):
i = 1
result = 0
# iterate up to sqrt(n)
while (i * i < n):
if (n % i == 0):
result += 2
i += 1
# if i^2 == n, then we count an extra divisor
if (i * i == n):
result += 1
return result
```

## Checking for prime numbers

Based on the same principles as the example above, you can check for primes by iterating over `i * i <= n`

. If `n % i == 0`

, then the number isn’t prime.

## Sieve of Eratosthenes

Can be used to find all prime numbers in the range to .

Here’s an implementation based on Codility’s notes.

Some quick reminders:

- Need a list of flags for each number from
`2`

to`n`

. - Main loop iterates from 2 to (inclusive). Not necessary to iterate all the way to
`n`

. - Each time the main loop is incremented, if the flag for the current value
`i`

indicates it is still prime:- Mark as non prime.
- Mark as non prime.
- Continue marking multiples as non prime until reaching
`n`

.

- At the end, all prime numbers will have been found.
- Complexity is .

### Finding prime factors

The Sieve of Eratosthenes algorithm can be modified to find the prime factors of a number `n`

.

- Build an array
`F`

of minimum prime factor for each value of`i`

instead of just using`true/false`

. - Iterate through this array starting at element
`n`

. - Store the prime factor at position
`F[n]`

in the output list of prime factors. `n = n / F[n]`

- Keep iterating until
`F[n] == 0`

Complexity is .

See here for an implementation based on Codility’s notes.

### Euclidean algorithm

The Codility notes provide 3 approaches for finding the greatest common divisor (**gcd**) between 2 numbers:

**Euclidean algorithm by subtraction:**recursively subtract the larger value from the smaller until the values are equal.**Euclidean algorithm by division:**recursively use the modulo operator until the two values are divisible by each other. where and are the two input values.**Binary Euclidean algorithm:**a more complex implementation involving division by 2 and scaling the result.

The least common multiple (**lcm**) of and is the smallest value that can be divided by and .

See here for an implementation based on Codility’s notes.

### Fibonacci numbers

```
def fib(n):
fib = [0] * (n + 2)
fib[1] = 1
for i in range(2, n + 1):
fib[i] = fib[n - 1] + fib[n - 2]
return fib[n]
```

Complexity is .

### Binary search

Start with a sorted list of values `v`

, length `n`

. `beg = 0`

, `end = n - 1`

, . Compare the value of `v[mid]`

to `x`

, and set `beg`

or `end`

to `mid ± 1`

depending on the value. Keep repeating until `x`

has been found (or can’t be found).

Complexity is .